This document is the copyright page and preface for the book "Applied Statistics and Probability for Engineers" by Douglas C. Montgomery and George C. Runger. The copyright is held by John Wiley & Sons, Inc. in 2003. This book was edited, designed, and produced by various teams at John Wiley & Sons and printed by Donnelley/Willard. The preface states that the purpose of the included Student Solutions Manual is to provide additional help for students in understanding the problem-solving processes presented in the main text.

1. Applied Statistics
and Probability
for Engineers
Third Edition
Douglas C. Montgomery
Arizona State University
George C. Runger
Arizona State University
John Wiley & Sons, Inc.

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Library of Congress Cataloging-in-Publication Data
Montgomery, Douglas C.
Applied statistics and probability for engineers / Douglas C. Montgomery, George C.
Runger.—3rd ed.
p. cm.
Includes bibliographical references and index.
ISBN 0-471-20454-4 (acid-free paper)
1. Statistics. 2. Probabilities. I. Runger, George C. II. Title.
QA276.12.M645 2002
519.5—dc21
2002016765
Printed in the United States of America.
10 9 8 7 6 5 4 3 2 1

3. SO fm.qxd 8/6/02 4:31 PM Page v
Preface
The purpose of this Student Solutions Manual is to provide you with additional help in under-
standing the problem-solving processes presented in Applied Statistics and Probability for
Engineers. The Applied Statistics text includes a section entitled “Answers to Selected
Exercises,” which contains the final answers to most odd-numbered exercises in the book.
Within the text, problems with an answer available are indicated by the exercise number
enclosed in a box.
This Student Solutions Manual provides complete worked-out solutions to a subset of the
problems included in the “Answers to Selected Exercises.” If you are having difficulty reach-
ing the final answer provided in the text, the complete solution will help you determine the
correct way to solve the problem.
Those problems with a complete solution available are indicated in the “Answers to
Selected Exercises,” again by a box around the exercise number. The complete solutions to
this subset of problems may also be accessed by going directly to this Student Solutions
Manual.

4. Chapter 2 Selected Problem Solutions
Section 2-2
2-43. 3 digits between 0 and 9, so the probability of any three numbers is 1/(10*10*10);
3 letters A to Z, so the probability of any three numbers is 1/(26*26*26); The probability your license plate
-8
is chosen is then (1/103)*(1/263) = 5.7 x 10
Section 2-3
2-49. a) P(A') = 1- P(A) = 0.7
b) P ( A ∪ B ) = P(A) + P(B) - P( A ∩ B ) = 0.3+0.2 - 0.1 = 0.4
c) P( A ′ ∩ B ) + P( A ∩ B ) = P(B). Therefore, P( A ′ ∩ B ) = 0.2 - 0.1 = 0.1
d) P(A) = P( A ∩ B ) + P( A ∩ B ′ ). Therefore, P( A ∩ B ′ ) = 0.3 - 0.1 = 0.2
e) P(( A ∪ B )') = 1 - P( A ∪ B ) = 1 - 0.4 = 0.6
f) P( A ′ ∪ B ) = P(A') + P(B) - P( A ′ ∩ B ) = 0.7 + 0.2 - 0.1 = 0.8
Section 2-4
2-61. Need data from example
a) P(A) = 0.05 + 0.10 = 0.15
P( A ∩ B) 0.04 + 0.07
b) P(A|B) = = = 0.153
P( B ) 0.72
c) P(B) = 0.72
P( A ∩ B) 0.04 + 0.07
d) P(B|A) = = = 0.733
P( B ) 0.15
e) P(A ∩ B) = 0.04 +0.07 = 0.11
f) P(A ∪ B) = 0.15 + 0.72 – 0.11 = 0.76
2-67. a) P(gas leak) = (55 + 32)/107 = 0.813
b) P(electric failure|gas leak) = (55/107)/(87/102) = 0.632
c) P(gas leak| electric failure) = (55/107)/(72/107) = 0.764
Section 2-5
2-73. Let F denote the event that a roll contains a flaw.
Let C denote the event that a roll is cotton.
P ( F ) = P( F C ) P(C ) + P( F C ′) P (C ′)
= (0.02)(0.70) + (0.03)(0.30) = 0.023
2-79. Let A denote a event that the first part selected has excessive shrinkage.
Let B denote the event that the second part selected has excessive shrinkage.
a) P(B)= P( B A )P(A) + P( B A ')P(A')
= (4/24)(5/25) + (5/24)(20/25) = 0.20
b) Let C denote the event that the second part selected has excessive shrinkage.
P(C) = P(C A ∩ B)P( A ∩ B) + P(C A ∩ B')P( A ∩ B')
+P(C A'∩B)P( A'∩B) + P(C A'∩B')P( A'∩B')
3  2   5  4  20   5  4  5   20  5  19   20 
=    +    +    +   
23  24   25  23  24   25  23  24   25  23  24   25 
= 0.20

5. Section 2-6
2-87. It is useful to work one of these exercises with care to illustrate the laws of probability. Let Hi denote the
event that the ith sample contains high levels of contamination.
' ' ' ' ' ' ' ' ' '
a) P(H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 ) = P(H1)P(H2 )P(H3 )P(H4 )P(H5 )
by independence. Also, P(Hi' ) = 0.9 . Therefore, the answer is 0.9 5 = 0.59
b) A1 = (H1 ∩ H'2 ∩ H3 ∩ H4 ∩ H5 )
' ' '
' ' ' '
A 2 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 )
' ' ' '
A 3 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 )
' ' ' '
A 4 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 )
' ' ' '
A 5 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 )
The requested probability is the probability of the union A1 ∪ A 2 ∪ A 3 ∪ A 4 ∪ A 5 and these events
are mutually exclusive. Also, by independence P( A i ) = 0.9 4 (0.1) = 0.0656 . Therefore, the answer is
5(0.0656) = 0.328.
c) Let B denote the event that no sample contains high levels of contamination. The requested
probability is P(B') = 1 - P(B). From part (a), P(B') = 1 - 0.59 = 0.41.
2-89. Let A denote the event that a sample is produced in cavity one of the mold.
1
a) By independence, P( A1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A 5 ) = ( )5 = 0.00003
8
b) Let Bi be the event that all five samples are produced in cavity i. Because the B's are mutually
exclusive, P(B1 ∪ B 2 ∪...∪B 8 ) = P(B1) + P(B 2 )+...+P(B 8 )
1 1
From part a., P(Bi ) = ( )5 . Therefore, the answer is 8( )5 = 0.00024
8 8
1 7
c) By independence, P( A 1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A '5 ) = ( ) 4 ( ) . The number of sequences in
8 8
1 7
which four out of five samples are from cavity one is 5. Therefore, the answer is 5( ) 4 ( ) = 0.00107 .
8 8
Section 2-7
2-97. Let G denote a product that received a good review. Let H, M, and P denote products that were high,
moderate, and poor performers, respectively.
a)
P(G ) = P(G H ) P( H ) + P(G M ) P( M ) + P(G P) P( P)
= 0. 95( 0. 40) + 0. 60( 0. 35) + 0. 10( 0. 25)
= 0. 615
b) Using the result from part a.,
P ( G H ) P ( H ) 0. 95( 0. 40)
P( H G ) = = = 0. 618
P( G ) 0. 615
c) P ( H G ' ) = P ( G ' H ) P ( H ) = 0. 05( 0. 40) = 0. 052
P(G' ) 1 − 0. 615
Supplemental
2-105. a) No, P(E1 ∩ E2 ∩ E3) ≠ 0
b) No, E1′ ∩ E2′ is not ∅
c) P(E1′ ∪ E2′ ∪ E3′) = P(E1′) + P(E2′) + P(E3′) – P(E1′∩ E2′) - P(E1′∩ E3′) - P(E2′∩ E3′)
+ P(E1′ ∩ E2′ ∩ E3′)
= 40/240

6. d) P(E1 ∩ E2 ∩ E3) = 200/240
e) P(E1 ∪ E3) = P(E1) + P(E3) – P(E1 ∩ E3) = 234/240
f) P(E1 ∪ E2 ∪ E3) = 1 – P(E1′ ∩ E2′ ∩ E3′) = 1 - 0 = 1
2-107. Let Ai denote the event that the ith bolt selected is not torqued to the proper limit.
a) Then,
P( A 1 ∩ A 2 ∩ A 3 ∩ A 4 ) = P( A 4 A 1 ∩ A 2 ∩ A 3 )P( A 1 ∩ A 2 ∩ A 3 )
= P( A 4 A 1 ∩ A 2 ∩ A 3 )P( A 3 A 1 ∩ A 2 )P( A 2 A 1)P( A 1)
 2  3  4  5 
=         = 0.282
 17   18   19   20 
b) Let B denote the event that at least one of the selected bolts are not properly torqued. Thus, B' is the
event that all bolts are properly torqued. Then,
 15   14   13   12 
P(B) = 1 - P(B') = 1 −         = 0.718
 20   19   18   17 
2-113. D = defective copy
 2  73  72   73  2  72   73  72  2 
a) P(D = 1) =     +     +     = 0.0778
 75  74  73   75  74  73   75  74  73 
 2  1  73   2  73  1   73  2  1 
b) P(D = 2) =     +     +     = 0.00108
 75  74  73   75  74  73   75  74  73 
2-117. Let A i denote the event that the ith washer selected is thicker than target.
 30  29  28 
a)     = 0.207
 50  49  8 
b) 30/48 = 0.625
c) The requested probability can be written in terms of whether or not the first and second washer selected
are thicker than the target. That is,
' ' ' '
P( A 3 ) = P( A 1A 2 A 3 orA 1A 2 A 3 orA 1A 2 A 3 orA 1A 2 A 3 )
' '
= P( A 3 A 1 A 2 )P( A 1A 2 ) + P( A 3 A 1 A 2 )P( A 1A 2 )
' ' ' ' '
+P( A 3 A 1 'A 2 )P( A 1A 2 ) + P( A 3 A 1 A 2 )P( A 1A 2 )
' '
= P( A 3 A 1 A 2 )P( A 2 A 1 )P( A 1 ) + P( A 3 A 1 A 2 )P( A 2 A 1 )P( A 1 )
' ' ' ' ' ' ' '
+P( A 3 A 1 A 2 )P( A 2 A 1 )P( A 1 ) + P( A 3 A 1 A 2 )P( A 2 A 1 )P( A 1 )
28  30 29  29  20 30  29  20 30  30  20 19 
=  +  +  +  
48  50 49  48  50 49  48  50 49  48  50 49 
= 0.60
2-121. Let A i denote the event that the ith row operates. Then,
P ( A1 ) = 0. 98, P ( A 2 ) = ( 0. 99)( 0. 99) = 0. 9801, P ( A 3 ) = 0. 9801, P ( A 4 ) = 0. 98.
The probability the circuit does not operate is
P( A1' ) P( A2 ) P( A3 ) P ( A4 ) = (0.02)(0.0199)(0.0199)(0.02) = 1.58 × 10 −7
' ' '

7. Chapter 3 Selected Problem Solutions
Section 3-2
3-13.
f X (0) = P( X = 0) = 1 / 6 + 1 / 6 = 1 / 3
f X (1.5) = P( X = 1.5) = 1 / 3
f X ( 2) = 1 / 6
f X (3) = 1 / 6
3-21. P(X = 0) = 0.023 = 8 x 10-6
P(X = 1) = 3[0.98(0.02)(0.02)]=0.0012
P(X = 2) = 3[0.98(0.98)(0.02)]=0.0576
P(X = 3) = 0.983 = 0.9412
3-25. X = number of components that meet specifications
P(X=0) = (0.05)(0.02)(0.01) = 0.00001
P(X=1) = (0.95)(0.02)(0.01) + (0.05)(0.98)(0.01)+(0.05)(0.02)(0.99) = 0.00167
P(X=2) = (0.95)(0.98)(0.01) + (0.95)(0.02)(0.99) + (0.05)(0.98)(0.99) = 0.07663
P(X=3) = (0.95)(0.98)(0.99) = 0.92169
Section 3-3
3-27.
 0, x < −2 
 
 1 / 8 −2 ≤ x < −1
3 / 8 −1 ≤ x < 0 
 
F( x) =  
5/8 0 ≤ x<1 
7 / 8 1 ≤ x < 2 
 

 1 2≤x  
a) P(X ≤ 1.25) = 7/8
b) P(X ≤ 2.2) = 1
c) P(-1.1 < X ≤ 1) = 7/8 − 1/8 = 3/4
d) P(X > 0) = 1 − P(X ≤ 0) = 1 − 5/8 = 3/8
3-31.
 0 x<0  .
0.008 0 ≤ x < 1 f (0) = 0.2 3 = 0.008,

 

F ( x) = 0.104 1 ≤ x < 2  where f (1) = 3(0.2)(0.2)(0.8) = 0.096,
0.488 2 ≤ x < 3 f (2) = 3(0.2)(0.8)(0.8) = 0.384,
 
 1
 3≤ x   f (3) = (0.8) 3 = 0.512,
3-33. a) P(X ≤ 3) = 1
b) P(X ≤ 2) = 0.5
c) P(1 ≤ X ≤ 2) = P(X=1) = 0.5
d) P(X>2) = 1 − P(X≤2) = 0.5

8. Section 3-4
3-37 Mean and Variance
µ = E ( X ) = 0 f (0) + 1 f (1) + 2 f (2) + 3 f (3) + 4 f (4)
= 0(0.2) + 1(0.2) + 2(0.2) + 3(0.2) + 4(0.2) = 2
V ( X ) = 0 2 f (0) + 12 f (1) + 2 2 f (2) + 3 2 f (3) + 4 2 f (4) − µ 2
= 0(0.2) + 1(0.2) + 4(0.2) + 9(0.2) + 16(0.2) − 2 2 = 2
3-41. Mean and variance for exercise 3-19
µ = E ( X ) = 10 f (10) + 5 f (5) + 1 f (1)
= 10(0.3) + 5(0.6) + 1(0.1)
= 6.1 million
V ( X ) = 10 2 f (10) + 5 2 f (5) + 12 f (1) − µ 2
= 10 2 (0.3) + 5 2 (0.6) + 12 (0.1) − 6.12
= 7.89 million 2
3-45. Determine x where range is [0,1,2,3,x] and mean is 6.
µ = E ( X ) = 6 = 0 f (0) + 1 f (1) + 2 f (2) + 3 f (3) + xf ( x)
6 = 0(0.2) + 1(0.2) + 2(0.2) + 3(0.2) + x(0.2)
6 = 1.2 + 0.2 x
4.8 = 0.2 x
x = 24
Section 3-5
3-47. E(X) = (3+1)/2 = 2, V(X) = [(3-1+1)2 -1]/12 = 0.667
3-49. X=(1/100)Y, Y = 15, 16, 17, 18, 19.
1  15 + 19 
E(X) = (1/100) E(Y) =   = 0.17 mm
100  2 
 1   (19 − 15 + 1) − 1
2 2
V (X ) =     = 0.0002 mm
2
 100   12 
Section 3-6
10 
3-57. a) P ( X = 5) =  0.5 5 (0.5) 5 = 0.2461
5
 
10  0 10 10  1 9 10  2 8
b) P ( X ≤ 2) =  0.5 0.5 +  0.5 0.5 +  0.5 0.5
0 1 2
     
= 0.510 + 10(0.5)10 + 45(0.5)10 = 0.0547

9. 10  10 
c) P( X ≥ 9) =  0.5 9 (0.5)1 +  0.510 (0.5) 0 = 0.0107
9 10 
   
10  10 
d) P(3 ≤ X < 5) =  0.530.57 +  0.54 0.56
3 4
   
= 120(0.5)10 + 210(0.5)10 = 0.3223
3-61. n=3 and p=0.25
3
3 27
f (0) =   =
 0 x<0  4 64
0.4219 0 ≤ x < 1
2
 1  3  27
  f (1) = 3   =
   4  4  64
F ( x) = 0.8438 1 ≤ x < 2  where 2
0.9844 2 ≤ x < 3 1 3 9
f (2) = 3    =
   4   4  64
 1
 3≤ x   3
1 1
f (3) =   =
4 64
1000 
3-63. = 1) = 
a) P ( X  1 0.001 (0.999) = 0.3681

1 999
 
1000 
 1 0.001 (0.999 ) = 0.6323
b) P( X ≥ 1) = 1 − P( X = 0) = 1 −  1 999

 
1000  1000 
c) P ( X ≤ 2) =  0 0.001 (0.999 ) +  1 0.001 (0.999 ) + ( 2 )0.001 0.999
0 1000 1 999 1000 2 998
  
   
= 0.9198
d ) E ( X ) = 1000(0.001) = 1
V ( X ) = 1000(0.001)(0.999) = 0.999
3-67. Let X denote the passengers with tickets that do not show up for the flight. Then, X is binomial
with n = 125 and p = 0.1.
a) P( X ≥ 5) = 1 − P( X ≤ 4)
125  0 125  1 125  2
 0 0.1 (0.9 ) +  1 0.1 (0.9 ) +  2 0.1 (0.9 )
= 1 − 
125 124 123
    
     
125  3 125  4 121 
+ 0.1 (0.9 ) +  4 0.1 (0.9 )  = 0.9961
122
 3  
    
b) P( X > 5) = 1 − P( X ≤ 5) = 0.9886

10. 3-69. Let X denote the number of questions answered correctly. Then, X is binomial with n = 25
and p = 0.25.
 25   25  25
a) P( X ≥ 20) =  0.25 20 (0.75) +  0.25 21 (0.75) +  0.25 22 (0.75)
 
5 4 3
 21  22 
 20     
 25  25   25
+  0.25 23 (0.75) +  0.25 24 (0.75) +  0.25 25 (0.75) ≅ 0
2 1 0
 23  24   25
     
 25  25  25
b) P( X < 5) =  0.25 0 (0.75) +  0.251 (0.75) +  0.25 2 (0.75)
 
25 24 23
1 2
0    
 25  25
+  0.253 (0.75) +  0.25 4 (0.75) = 0.2137
22 21
3 4
   
Section 3-7
3-71. a. P( X= 1) = (1 − 0.5) 0 0.5 = 0.5
b. P( X= 4) = (1 − 0.5) 3 0.5 = 0.5 4 = 0.0625
c. P( X= 8) = (1 − 0.5) 7 0.5 = 0.58 = 0.0039
d. P( X≤ 2) = P ( X = 1) + P( X = 2) = (1 − 0.5) 0 0.5 + (1 − 0.5)1 0.5
= 0.5 + 0.5 2 = 0.75
e. P ( X > 2) = 1 − P ( X ≤ 2) = 1 − 0.75 = 0.25
3-75. Let X denote the number of calls needed to obtain a connection. Then, X is a geometric random variable
with p = 0.02
a) P( X = 10) = (1 − 0.02) 9 0.02 = 0.989 0.02 = 0.0167
b) P( X > 5) = 1 − P( X ≤ 4) = 1 − [ P( X = 1) + P( X = 2) + P( X = 3) + P( X = 4)]
= 1 − [0.02 + 0.98(0.02) + 0.98 2 (0.02) + 0.98 3 (0.02)]
= 1 − 0.0776 = 0.9224
c) E(X) = 1/0.02 = 50
3-77 p = 0.005 , r = 8
a. P( X = 8) = 0.0005 8 = 3.91x10 −19
1
b. µ = E( X ) = = 200 days
0.005
c Mean number of days until all 8 computers fail. Now we use p=3.91x10-19
1
µ = E (Y ) = −91
= 2.56 x1018 days or 7.01 x1015 years
3.91x10
3-81. a) E(X) = 4/0.2 = 20
19 
b) P(X=20) =  (0.80)16 0.24 = 0.0436
3
 
18 
c) P(X=19) =  (0.80) 0.2 = 0.0459
15 4
3
 
 20 
d) P(X=21) = 
 3 (0.80) 0.2 = 0.0411
17 4

 

11. e) The most likely value for X should be near µX. By trying several cases, the most likely value is x = 19.
3-83. Let X denote the number of fills needed to detect three underweight packages. Then X is a negative
binomial random variable with p = 0.001 and r = 3.
a) E(X) = 3/0.001 = 3000
b) V(X) = [3(0.999)/0.0012] = 2997000. Therefore, σX = 1731.18
Section 3-8
P( X = 1) =
( )( ) = (4 ×16 ×15 ×14) / 6 = 0.4623
4
1
16
3
( ) (20 ×19 ×18 ×17) / 24
3-87. a)
20
4
b) P ( X = 4) =
( )( ) =
4 16
4 0 1
= 0.00021
( ) (20 ×19 ×18 ×17) / 24
20
4
c)
P( X ≤ 2) = P( X = 0) + P( X = 1) + P( X = 2)
=
( )( ) + ( )( ) + ( )( )
4
0
16
4
4
1
16
3
4 16
( ) ( ) ( )
2 2
20 20 20
4 4 4
 16×15×14×13 4×16×15×14 6×16×15 
 + + 
=  24

6
20×19×18×17 
2 
= 0.9866
 
 24 
d) E(X) = 4(4/20) = 0.8
V(X) = 4(0.2)(0.8)(16/19) = 0.539
3-91. Let X denote the number of men who carry the marker on the male chromosome for an increased risk for high
blood pressure. N=800, K=240 n=10
a) n=10
P( X = 1) =
( )( ) = ( )( ) = 0.1201
240 560
1 9
240! 560!
1!239! 9!551!
( ) 800
10
800!
10!790!
b) n=10
P( X > 1) = 1 − P( X ≤ 1) = 1 − [ P( X = 0) + P( X = 1)]
P( X = 0) =
( )( ) = (
240 560
0 10
240!
)( 560!
0!240! 10!560! )
= 0.0276
( ) 800
10
800!
10!790!
P( X > 1) = 1 − P( X ≤ 1) = 1 − [0.0276 + 0.1201] = 0.8523
Section 3-9
e −4 4 0
3-97. a) P( X = 0) = = e −4 = 0.0183
0!
b) P( X ≤ 2) = P( X = 0) + P ( X = 1) + P ( X = 2)
−4 e − 4 41 e − 4 4 2
=e + + = 0.2381
1! 2!
e −4 4 4
c) P( X = 4) = = 0.1954
4!

12. e −4 4 8
d) P( X = 8) = = 0.0298
8!
3-99. P( X = 0) = e − λ = 0.05 . Therefore, λ = −ln(0.05) = 2.996.
Consequently, E(X) = V(X) = 2.996.
3-101. a) Let X denote the number of flaws in one square meter of cloth. Then, X is a Poisson random variable
e −0.1 (0.1) 2
with λ = 0.1. P( X = 2) = = 0.0045
2!
b) Let Y denote the number of flaws in 10 square meters of cloth. Then, Y is a Poisson random variable
e −111
with λ = 1. P(Y = 1) = = e −1 = 0.3679
1!
c) Let W denote the number of flaws in 20 square meters of cloth. Then, W is a Poisson random variable
P(W = 0) = e −2 = 0.1353
with λ = 2.
d) P(Y ≥ 2) = 1 − P(Y ≤ 1) = 1 − P(Y = 0) − P(Y = 1)
= 1 − e −1 − e −1
= 0.2642
3-105. a) Let X denote the number of flaws in 10 square feet of plastic panel. Then, X is a Poisson random
−0.5
variable with λ = 0.5. P ( X = 0) = e = 0.6065
b) Let Y denote the number of cars with no flaws,
 10 
P (Y = 10 ) =   ( 0 . 3935 ) 10 ( 0 . 6065 ) 0 = 8 . 9 x10 − 5
 10 
 
c) Let W denote the number of cars with surface flaws. Because the number of flaws has a
Poisson distribution, the occurrences of surface flaws in cars are independent events with
constant probability. From part a., the probability a car contains surface flaws is 1−0.6065 =
0.3935. Consequently, W is binomial with n = 10 and p = 0.3935.
10 
P (W = 0) =  (0.6065) 0 (0.3935)10 = 8.9 x10 −5
0
 
10 
P (W = 1) =  (0.6065)1 (0.3935) 9 = 0.001372
 
1
P (W ≤ 1) = 0.000089 + 0.001372 = 0.00146
Supplemental Exercises
3-107. Let X denote the number of totes in the sample that do not conform to purity requirements. Then, X has a
hypergeometric distribution with N = 15, n = 3, and K = 2.
 2 13 
  
 0  3 
P( X ≥ 1) = 1 − P( X = 0) = 1 −    = 1 −
13!12!
= 0.3714
15  10! !
15
 
3
 

13. 3-109. Let Y denote the number of calls needed to obtain an answer in less than 30 seconds.
a) P (Y = 4) = (1 − 0.75) 0.75 = 0.25 3 0.75 = 0.0117
3
b) E(Y) = 1/p = 1/0.75 = 1.3333
e −5 5 5
3-111. a) Let X denote the number of messages sent in one hour. P( X = 5) = = 0.1755
5!
b) Let Y denote the number of messages sent in 1.5 hours. Then, Y is a Poisson random variable with
e −7.5 (7.5)10
λ =7.5. P(Y = 10) = = 0.0858
10!
c) Let W denote the number of messages sent in one-half hour. Then, W is a Poisson random variable with
λ = 2.5. P (W < 2) = P (W = 0) + P (W = 1) = 0.2873
3-119. Let X denote the number of products that fail during the warranty period. Assume the units are
independent. Then, X is a binomial random variable with n = 500 and p = 0.02.
 500 
a) P(X = 0) = 
 0 (0.02) (0.98) = 4.1 x 10

0 500 -5
 
b) E(X) = 500(0.02) = 10
c) P(X >2) = 1 − P(X ≤ 1) = 0.9995
3-121. a) P(X ≤ 3) = 0.2 + 0.4 = 0.6
b) P(X > 2.5) = 0.4 + 0.3 + 0.1 = 0.8
c) P(2.7 < X < 5.1) = 0.4 + 0.3 = 0.7
d) E(X) = 2(0.2) + 3(0.4) + 5(0.3) + 8(0.1) = 3.9
e) V(X) = 22(0.2) + 32(0.4) + 52(0.3) + 82(0.1) − (3.9)2 = 3.09
3-125. Let X denote the number of orders placed in a week in a city of 800,000 people. Then X is a Poisson
random variable with λ = 0.25(8) = 2.
a) P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − [e-2 + e-2(2) + (e-222)/2!] = 1 − 0.6767 = 0.3233.
b) Let Y denote the number of orders in 2 weeks. Then, Y is a Poisson random variable with λ = 4, and
P(Y<2) = P(Y ≤ 1) = e-4 + (e-441)/1! = 0.0916.
3-127. Let X denote the number of totes in the sample that exceed the moisture content. Then X is a binomial
random variable with n = 30. We are to determine p.
 30  0
If P(X ≥ 1) = 0.9, then P(X = 0) = 0.1. Then  ( p) (1 − p )30 = 0.1 , giving 30ln(1−p)=ln(0.1),
0
 
which results in p = 0.0738.
3-129. a) Let X denote the number of flaws in 50 panels. Then, X is a Poisson random variable with
λ = 50(0.02) = 1. P(X = 0) = e-1 = 0.3679.
b) Let Y denote the number of flaws in one panel, then
P(Y ≥ 1) = 1 − P(Y=0) = 1 − e-0.02 = 0.0198. Let W denote the number of panels that need to be
inspected before a flaw is found. Then W is a geometric random variable with p = 0.0198 and
E(W) = 1/0.0198 = 50.51 panels.
−0.02
c.) P (Y ≥ 1) = 1 − P (Y = 0) = 1 − e = 0.0198
Let V denote the number of panels with 2 or more flaws. Then V is a binomial random
variable with n=50 and p=0.0198

14.  50   50 
P(V ≤ 2) =  0.0198 0 (.9802) 50 +  0.01981 (0.9802) 49
0  
  1
 50 
+  0.0198 2 (0.9802) 48 = 0.9234
2
 

15. Chapter 4 Selected Problem Solutions
Section 4-2
∞
∞
4-1. ∫
a) P (1 < X ) = e − x dx = (−e − x )
1
1
= e −1 = 0.3679
2.5
2.5
∫e
−x
b) P (1 < X < 2.5) = dx = (−e− x ) = e −1 − e− 2.5 = 0.2858
1
1
3
∫
c) P ( X = 3) = e − x dx = 0
3
4
4
∫
d) P ( X < 4) = e − x dx = (−e − x ) = 1 − e − = 0.9817
4
0
0
∞
∞
∫
e) P (3 ≤ X ) = e− x dx = (−e − x )
3
3
= e − 3 = 0.0498
4 4
x x2 4 2 − 32
4-3 a) P ( X < 4) = ∫ dx = = = 0.4375 , because f X ( x) = 0 for x < 3.
3
8 16 3
16
5 5
x x2 5 2 − 3.5 2
b) , P ( X > 3.5) = ∫ dx = = = 0.7969 because f X ( x) = 0 for x > 5.
3. 5
8 16 3.5
16
5 2 5
x x 52 − 4 2
c) P (4 < X < 5) = ∫ 8 dx = 16
4
=
16
= 0.5625
4
4.5 2 4.5
x x 4.5 2 − 3 2
d) P ( X < 4.5) = ∫ 8 dx = 16
3
=
16
= 0.7031
3
5 3.5 5 3.5
x x x2 x2 52 − 4.52 3.52 − 32
e) P( X > 4.5) + P( X < 3.5) = ∫ dx + ∫ dx = + = + = 0.5 .
4.5 8 3 8 16 4.5 16 3 16 16
4-9 a) P(X < 2.25 or X > 2.75) = P(X < 2.25) + P(X > 2.75) because the two events are
mutually exclusive. Then, P(X < 2.25) = 0 and
2.8
P(X > 2.75) = ∫ 2dx = 2(0.05) = 0.10 .
2.75
b) If the probability density function is centered at 2.5 meters, then f X ( x) = 2 for
2.25 < x < 2.75 and all rods will meet specifications.
Section 4-3
4-11. a) P(X<2.8) = P(X ≤ 2.8) because X is a continuous random variable.
Then, P(X<2.8) =F(2.8)=0.2(2.8) = 0.56.
b) P ( X > 1.5) = 1 − P ( X ≤ 1.5) = 1 − 0.2(1.5) = 0.7

16. c) P ( X < −2) = FX (−2) = 0
d) P ( X > 6) = 1 − FX (6) = 0
x
x
−x
4-13. Now, f X ( x ) = e ∫
for 0 < x and F X ( x) = e − x dx = − e − x
0
0
= 1− e−x
 0, x ≤ 0
for 0 < x. Then, FX ( x) =  −x
1 − e , x > 0
x
x
4-21. F ( x) = ∫ 0.5 xdx = 0.5 x 2
2 = 0.25 x 2 for 0 < x < 2. Then,
0
0
0, x<0



F ( x) = 0.25 x 2 , 0≤ x<2


1, 2≤x

Section 4-4
5 5
x x3 53 − 33
4-25. E ( X ) = ∫ x dx = = = 4.083
3
8 24 3 24
5 5
x
V ( X ) = ∫ ( x − 4.083) 2 dx = ∫ x 2 8 dx − 4.083 2
x
3
8 3
5
x4
= − 4.083 2 = 0.3291
32 3
120
600 120
4-27. a.) E ( X ) = ∫x
100 x 2
dx = 600 ln x 100 = 109.39
120 120
600 (109.39 ) 2
V ( X ) = ∫ ( x − 109.39) 2 dx = 600 ∫ 1 − 2 (109.39 )
x
+ x2
dx
100 x2 100
120
= 600( x − 218.78 ln x − 109.39 2 x −1 ) 100
= 33.19
b.) Average cost per part = $0.50*109.39 = $54.70

17. Section 4-5
4-33. a) f(x)= 2.0 for 49.75 < x < 50.25.
E(X) = (50.25 + 49.75)/2 = 50.0,
(50.25 − 49.75) 2
V (X ) = = 0.0208, and σ x = 0.144 .
12
x
b) F ( x) = ∫ 2.0dx for 49.75 < x < 50.25.
49.75
Therefore,
0, x < 49.75



F ( x) = 2 x − 99.5, 49.75 ≤ x < 50.25


1, 50.25 ≤ x

c) P ( X < 50.1) = F (50.1) = 2(50.1) − 99.5 = 0.7
(1.5 + 2.2)
4-35 E( X ) = = 1.85 min
2
(2.2 − 1.5) 2
V (X ) = = 0.0408 min 2
12
2 2
1 2
b) P ( X < 2) = ∫ dx = ∫ 0.7 dx = 0.7 x 1.5 = 0.7(.5) = 0.7143
1. 5
(2.2 − 1.5) 1.5
x x
1 x
c.) F ( X ) = ∫5 (2.2 − 1.5) dx = 1∫50.7dx = 0.7 x 1.5
1. .
for 1.5 < x < 2.2. Therefore,
 0, x < 1.5

F ( x) = 0.7 x − 2.14, 1.5 ≤ x < 2.2
 1, 2.2 ≤ x

Section 4-6
4-41 a) P(Z < 1.28) = 0.90
b) P(Z < 0) = 0.5
c) If P(Z > z) = 0.1, then P(Z < z) = 0.90 and z = 1.28
d) If P(Z > z) = 0.9, then P(Z < z) = 0.10 and z = −1.28
e) P(−1.24 < Z < z) = P(Z < z) − P(Z < −1.24)
= P(Z < z) − 0.10749.
Therefore, P(Z < z) = 0.8 + 0.10749 = 0.90749 and z = 1.33

18. 4-43. a) P(X < 13) = P(Z < (13−10)/2)
= P(Z < 1.5)
= 0.93319
b) P(X > 9) = 1 − P(X < 9)
= 1 − P(Z < (9−10)/2)
= 1 − P(Z < −0.5)
= 1 − [1− P(Z < 0.5)]
= P(Z < 0.5)
= 0.69146.
6 − 10 14 − 10 
c) P(6 < X < 14) = P
 <Z< 
 2 2 
= P(−2 < Z < 2)
= P(Z < 2) −P(Z < − 2)]
= 0.9545.
2 − 10 4 − 10 
d) P(2 < X < 4) = P
 <Z< 
 2 2 
= P(−4 < Z < −3)
= P(Z < −3) − P(Z < −4)
= 0.00135
e) P(−2 < X < 8) = P(X < 8) − P(X < −2)
8 − 10  −2 − 10 
= P Z <


 − P Z < 
 2   2 
= P(Z < −1) − P(Z < −6)
= 0.15866.
 45 − 65 
4-51. a) P(X <45) = P Z < 
 5 
= P(Z < -3)
= 0.00135
 65 − 60 
b) P(X > 65) = P Z > 
 5 
= PZ >1)
= 1- P(Z < 1)
= 1 - 0.841345
= 0.158655
 x − 60 
c) P(X < x) = P Z <  = 0.99.
 5 
x − 60
Therefore, 5 = 2.33 and x = 71.6
4-55. a) P(X > 90.3) + P(X < 89.7)
 90.3 − 90.2   89.7 − 90.2 
= P Z >  + P Z < 
 0.1   0.1 
= P(Z > 1) + P(Z < −5)
= 1 − P(Z < 1) + P(Z < −5)

19. =1 − 0.84134 +0
= 0.15866.
Therefore, the answer is 0.15866.
b) The process mean should be set at the center of the specifications; that is, at µ = 90.0.
 89.7 − 90 90.3 − 90 
c) P(89.7 < X < 90.3) = P <Z< 
 0.1 0.1 
= P(−3 < Z < 3) = 0.9973.
The yield is 100*0.9973 = 99.73%
 0.0026 − 0.002 
4-59. a) P(X > 0.0026) = P Z > 
 0.0004 
= P(Z > 1.5)
= 1-P(Z < 1.5)
= 0.06681.
 0.0014 − 0.002 0.0026 − 0.002 
b) P(0.0014 < X < 0.0026) = P <Z< 
 0.0004 0.0004 
= P(−1.5 < Z < 1.5)
= 0.86638.
 0.0014 − 0.002 0.0026 − 0.002 
c) P(0.0014 < X < 0.0026) = P <Z< 
 σ σ 
 − 0.0006 0.0006 
= P <Z< .
 σ σ 
0.0006 
Therefore, P Z <
  = 0.9975. Therefore,
0 . 0006 = 2.81 and σ = 0.000214.
σ
 σ 
Section 4-7
4-67 Let X denote the number of errors on a web site. Then, X is a binomial random variable
with p = 0.05 and n = 100. Also, E(X) = 100 (0.05) = 5 and
V(X) = 100(0.05)(0.95) = 4.75
 1− 5 
P( X ≥ 1) ≅ P Z ≥
  = P( Z ≥ −1.84) = 1 − P( Z < −1.84) = 1 − 0.03288 = 0.96712

 4.75 
4-69 Let X denote the number of hits to a web site. Then, X is a Poisson random variable with
a of mean 10,000 per day. E(X) = λ = 10,000 and V(X) = 10,000
a)
 10,200 − 10,000 
P( X ≥ 10,200) ≅ P Z ≥

 = P ( Z ≥ 2) = 1 − P( Z < 2)

 10,000 
= 1 − 0.9772 = 0.0228

20. Expected value of hits days with more than 10,200 hits per day is
(0.0228)*365=8.32 days per year
b.) Let Y denote the number of days per year with over 10,200 hits to a web site.
Then, Y is a binomial random variable with n=365 and p=0.0228.
E(Y) = 8.32 and V(Y) = 365(0.0228)(0.9772)=8.13
 15 − 8.32 
P(Y > 15) ≅ P Z ≥  = P( Z ≥ 2.34) = 1 − P( Z < 2.34)
 8.13 
= 1 − 0.9904 = 0.0096
Section 4-9
4-77. Let X denote the time until the first call. Then, X is exponential and
λ = E (1X ) = 15 calls/minute.
1
∞ x x ∞
− −
a) P ( X > 30) = ∫
30
1
15
e 15 dx = − e 15
30
= e − 2 = 0.1353
b) The probability of at least one call in a 10-minute interval equals one minus the
probability of zero calls in
a 10-minute interval and that is P(X > 10).
− x ∞
P( X > 10) = − e 15
= e − 2 / 3 = 0.5134 .
10
Therefore, the answer is 1- 0.5134 = 0.4866. Alternatively, the requested probability is
equal to P(X < 10) = 0.4866.
x 10
− 15
c) P (5 < X < 10) = − e = e −1 / 3 − e − 2 / 3 = 0.2031
5
− t x
d) P(X < x) = 0.90 and P ( X < x) = − e 15
= 1 − e − x / 15 = 0.90 . Therefore, x = 34.54
0
minutes.
4-79. Let X denote the time to failure (in hours) of fans in a personal computer. Then, X is an
exponential random variable and λ = 1 / E ( X ) = 0.0003 .
∞ ∞
∫ 0.0003e
− x 0.0003 − x 0.0003
a) P(X > 10,000) = dx = − e = e −3 = 0.0498
10 , 000 10 , 000
7 , 000 7 , 000
∫ 0.0003e
− x 0.0003
b) P(X < 7,000) = dx = − e − x 0.0003 = 1 − e − 2.1 = 0.8775
0 0
4-81. Let X denote the time until the arrival of a taxi. Then, X is an exponential random
variable with λ = 1 / E ( X ) = 0.1 arrivals/ minute.

21. ∞ ∞
∫ 0.1e dx = − e
− 0. 1 x − 0.1 x
a) P(X > 60) = = e − 6 = 0.0025
60 60
10 10
∫ 0.1e dx = − e
− 0.1 x − 0.1 x
b) P(X < 10) = = 1 − e −1 = 0.6321
0 0
4-83. Let X denote the distance between major cracks. Then, X is an exponential random
variable with λ = 1 / E ( X ) = 0.2 cracks/mile.
∞ ∞
∫ 0.2e dx = − e
− 0. 2 x − 0.2 x
a) P(X > 10) = = e − 2 = 0.1353
10 10
b) Let Y denote the number of cracks in 10 miles of highway. Because the distance
between cracks is exponential, Y is a Poisson random variable with λ = 10(0.2) = 2
cracks per 10 miles.
e −2 2 2
P(Y = 2) = = 0.2707
2!
c) σ X = 1 / λ = 5 miles.
4-87. Let X denote the number of calls in 3 hours. Because the time between calls is an
exponential random variable, the number of calls in 3 hours is a Poisson random variable.
Now, the mean time between calls is 0.5 hours and λ = 1 / 0.5 = 2 calls per hour = 6 calls in
3 hours.
P( X ≥ 4) = 1 − P( X ≤ 3) = 1 −  e 6 + e 6 + e 6 + e 6  = 0.8488
−6 0 −6 1 −6 2 −6 3
 0!
 1! 2! 3! 
Section 4-10
4-97. Let Y denote the number of calls in one minute. Then, Y is a Poisson random variable
with λ = 5 calls per minute.
e −5 5 4
a) P(Y = 4) = = 0.1755
4!
e −5 50 e −5 51 e −5 52
b) P(Y > 2) = 1 - P (Y ≤ 2) = 1 − − − = 0.8754 .
0! 1! 2!
Let W denote the number of one minute intervals out of 10 that contain more than 2
calls. Because the calls are a Poisson process, W is a binomial random variable with n =
10 and p = 0.8754.
( )
10 10 0
Therefore, P(W = 10) = 10 0.8754 (1 − 0.8754) = 0.2643 .
4-101. Let X denote the number of bits until five errors occur. Then, X has an Erlang distribution
with r = 5 and λ = 10 −5 error per bit.

22. r
a) E(X) = = 5 × 105 bits.
λ
r
b) V(X) = 2 = 5 × 10 and σ X = 5 × 10 = 223607 bits.
10 10
λ
c) Let Y denote the number of errors in 105 bits. Then, Y is a Poisson random variable
with
λ = 1 / 105 = 10−5 error per bit = 1 error per 105 bits.
−1 0
[
P(Y ≥ 3) = 1 − P(Y ≤ 2) = 1 − e 0!1 + e 1!1 + e 2!1 = 0.0803
−1 1 −1 2
]
4-105. a) Γ(6) = 5!= 120
b) Γ( 5 ) = 3 Γ( 3 ) =
2 2 2
3 1
2 2
Γ( 1 ) = 3 π 1 / 2 = 1.32934
2 4
c) Γ( 9 ) = 7 Γ( 7 ) = 7 5 3 1 Γ( 1 ) = 105 π 1 / 2 = 11.6317
2 2 2 2 2 2 2 2 16
Section 4-11
4-109. β=0.2 and δ=100 hours
E ( X ) = 100Γ(1 + 1
0. 2
) = 100 × 5!= 12,000
V ( X ) = 100 2 Γ(1 + 2
0.2
) − 100 2 [Γ(1 + 1
0.2
)]2 = 3.61 × 1010
4-111. Let X denote lifetime of a bearing. β=2 and δ=10000 hours
2
 8000 
−  2
a) P( X > 8000) = 1 − FX (8000) = e  10000 
= e − 0.8 = 0.5273
b)
E ( X ) = 10000Γ(1 + 1 ) = 10000Γ(1.5)
2
= 10000(0.5)Γ(0.5) = 5000 π = 8862.3
= 8862.3
hours
c) Let Y denote the number of bearings out of 10 that last at least 8000 hours. Then, Y is
a
binomial random variable with n = 10 and p = 0.5273.
( )
P(Y = 10) = 10 0.527310 (1 − 0.5273) 0 = 0.00166 .
10
Section 4-12
4-117 X is a lognormal distribution with θ=5 and ω2=9
a. )
 ln(13330) − 5 
P( X < 13300) = P(e W < 13300) = P (W < ln(13300)) = Φ 
 3 
= Φ(1.50) = 0.9332
b.) Find the value for which P(X≤x)=0.95

23.  ln( x) − 5 
P( X ≤ x) = P(e W ≤ x) = P(W < ln( x)) = Φ  = 0.95
 3 
ln( x) − 5
= 1.65 x = e 1.65(3) + 5 = 20952.2
3
2
c.) µ = E ( X ) = e θ +ω = e 5+ 9 / 2 = e 9.5 = 13359.7
/2
2 2
V ( X ) = e 2θ +ω (e ω − 1) = e 10 + 9 (e 9 − 1) = e 19 (e 9 − 1) = 1.45 × 1012
4-119 a.) X is a lognormal distribution with θ=2 and ω2=4
 ln(500) − 2 
P( X < 500) = P(e W < 500) = P(W < ln(500)) = Φ 
 2 
= Φ(2.11) = 0.9826
b.)
P(1000 < X < 1500)
P( X < 15000 | X > 1000) =
P( X > 1000)
  ln(1500) − 2   ln(1000) − 2 
Φ  2
 − Φ
2

    
=
  ln(1000) − 2 
1 − Φ 2

  
Φ(2.66) − Φ(2.45) 0.9961 − 0.9929
= = = 0.0032 / 0.007 = 0.45
(1 − Φ(2.45) ) (1 − 0.9929)
c.) The product has degraded over the first 1000 hours, so the probability of it lasting
another 500 hours is very low.
4-121 Find the values of θand ω2 given that E(X) = 100 and V(X) = 85,000

24. 100 2 2
x =
y 85000 = e 2θ +ω (eω − 1)
2
let x = eθ and y = e ω
2 2 2 2
then (1) 100 = x y and (2) 85000= x y( y −1) = x y − x y
2
Square (1) 10000 = x y and substitute into (2)
85000 = 10000 ( y − 1)
y = 9 .5
100
Substitute y into (1) and solve for x x = = 32.444
9.5
θ = ln(32.444) = 3.45 and ω 2 = ln(9.5) = 2.25
Supplemental Exercises
4-127. Let X denote the time between calls. Then, λ = 1 / E ( X ) = 0.1 calls per minute.
5 5
∫
a) P ( X < 5) = 0.1e − x dx = −e − x = 1 − e − 0.5 = 0.3935
0.1 0.1
0 0
15
b) P (5 < X < 15) = −e − 0.1 x = e − 0.5 − e −1.5 = 0.3834
5
x
∫
c) P(X < x) = 0.9. Then, P ( X < x) = 0.1e − t dt = 1 − e − x = 0.9 . Now, x = 23.03
0.1 0.1
0
minutes.
4-129. a) Let Y denote the number of calls in 30 minutes. Then, Y is a Poisson random variable
e −3 3 0 e −3 31 e −3 3 2
with x = e θ . P(Y ≤ 2) = + + = 0.423 .
0! 1! 2!
b) Let W denote the time until the fifth call. Then, W has an Erlang distribution with
λ = 0.1 and r = 5.
E(W) = 5/0.1 = 50 minutes
4-137. Let X denote the thickness.
5.5 − 5 

a) P(X > 5.5) = P Z >  = P(Z > 2.5) = 0. 0062
 0.2 
 4.5 − 5 5.5 − 5 
b) P(4.5 < X < 5.5) = P <Z <  = P (-2.5 < Z < 2.5) = 0.9876
 0.2 0.2 
Therefore, the proportion that do not meet specifications is 1 − P(4.5 < X < 5.5) =
0.012.

25.  x −5 x −5
c) If P(X < x) = 0.90, then P Z >  = 0.9. Therefore, = 1.65 and x = 5.33.
 0.2  0.2
4-139. If P(0.002-x < X < 0.002+x), then P(-x/0.0004 < Z < x/0.0004) = 0.9973. Therefore,
x/0.0004 = 3 and x = 0.0012. The specifications are from 0.0008 to 0.0032.
10 , 000 − µ 10, 000− µ
4-141. If P(X > 10,000) = 0.99, then P(Z > 600 ) = 0.99. Therefore, 600 = -2.33 and
µ = 11,398 .
4-143 X is an exponential distribution with E(X) = 7000 hours
5800 x  5800 
1 − − 
a.) P ( X < 5800) = ∫
0 7000
e 7000 dx = 1 − e  7000  = 0.5633
∞ x x
1 − −
b.) P ( X > x ) = ∫ e 7000 dx =0.9 Therefore, e 7000 = 0.9
x
7000
and x = −7000 ln(0.9) = 737.5 hours

26. Chapter 5 Selected Problem Solutions
Section 5-1
5-7.
E ( X ) = 1[ f XY (1,1) + f XY (1,2) + f XY (1,3)] + 2[ f XY (2,1) + f XY (2,2) + f XY (2,3)]
+ 3[ f XY (3,1) + f XY (3,2) + f XY (3,3)]
= (1 × 36 ) + (2 × 12 ) + (3 × 15 ) = 13 / 6 = 2.167
9
36 36
V ( X ) = (1 − 13 ) 2
6
9
36 + (2 − 13 ) 2
6
12
36 + (3 − 13 ) 2
6
15
36 = 0.639
E (Y ) = 2.167
V (Y ) = 0.639
5-11.
E ( X ) = −1( 1 ) − 0.5( 1 ) + 0.5( 1 ) + 1( 1 ) =
8 4 2 8
1
8
E (Y ) = −2( 1 ) − 1( 1 ) + 1( 1 ) + 2( 1 ) =
8 4 2 8
1
4
5-15 a) The range of (X,Y) is X ≥ 0, Y ≥ 0 and X + Y ≤ 4 . X is the number of pages with moderate
graphic content and Y is the number of pages with high graphic output out of 4.
x=0 x=1 x=2 x=3 x=4
-05
y=4 5.35x10 0 0 0 0
y=3 0.00183 0.00092 0 0 0
y=2 0.02033 0.02066 0.00499 0 0
y=1 0.08727 0.13542 0.06656 0.01035 0
y=0 0.12436 0.26181 0.19635 0.06212 0.00699
b.)
x=0 x=1 x=2 x=3 x=4
f(x) 0.2338 0.4188 0.2679 0.0725 0.0070
c.)
E(X)=
4
∑x 0
i f ( xi ) = 0(0.2338) + 1(0.4188) + 2(0.2679) + 3(0.7248) = 4(0.0070) = 1.2
f XY (3, y )
d.) f Y 3 ( y) = , fx(3) = 0.0725
f X (3)
y fY|3(y)
0 0.857
1 0.143
2 0
3 0
4 0
e) E(Y|X=3) = 0(0.857)+1(0.143) = 0.143

27. Section 5-2
5-17. a) P( X = 2) = f XYZ (2,1,1) + f XYZ (2,1,2) + f XYZ (2,2,1) + f XYZ (2,2,2) = 0.5
b) P( X = 1, Y = 2) = f XYZ (1,2,1) + f XYZ (1,2,2) = 0.35
c) P( Z < 1.5) = f XYZ (1,1,1) + f XYZ (1,2,2) + f XYZ (2,1,1) + f XYZ (2,2,1) = 0.5
d)
P( X = 1 or Z = 1) = P( X = 1) + P( Z = 1) − P( X = 1, Z = 1) = 0.5 + 0.5 − 0.2 = 0.8
e) E(X) = 1(0.5) + 2(0.5) = 1.5
5-25. P(X=x, Y=y, Z=z) is the number of subsets of size 4 that contain x printers with graphics enhancements, y
printers with extra memory, and z printers with both features divided by the number of subsets of size 4.
From the results on the CD material on counting techniques, it can be shown that
( )( )( )
4
x
5
y
6
z
P ( X = x, Y = y , Z = z ) =
( ) 15
for x+y+z = 4.
4
a) P ( X = 1, Y = 2, Z = 1) =
( )( )( ) = 0.1758
4 5
1 2
6
1
( ) 15
4
b) P ( X = 1, Y = 1) = P ( X = 1, Y = 1, Z = 2) =
( )( )( ) = 0.2198
4
1 1
5 6
2
( ) 15
4
c) The marginal distribution of X is hypergeometric with N = 15, n = 4, K = 4.
Therefore, E(X) = nK/N = 16/15 and V(X) = 4(4/15)(11/15)[11/14] = 0.6146.
P( X = 2, Y = 2) 0.1944
5-29 a.) P( X = 2 | Y = 2) = = = 0.7347
P(Y = 2) 0.2646
P( X = 2, Y = 2) = 0.1922
 4
P(Y = 2) =  0.3 2 0.7 4 = 0.2646
 2 from the binomial marginal distribution of Y
 
b) Not possible, x+y+z=4, the probability is zero.
c.) P( X | Y = 2) = P( X = 0 | Y = 2), P( X = 1 | Y = 2), P( X = 2 | Y = 2)
P( X = 0, Y = 2)  4! 
P( X = 0 | Y = 2) = = 0.6 0 0.3 2 0.12  0.2646 = 0.0204
P(Y = 2)  0!2!2! 
P ( X = 1, Y = 2)  4! 
P ( X = 1 | Y = 2) = = 0.610.3 2 0.11  0.2646 = 0.2449
P(Y = 2)  1!2!1! 
P( X = 2, Y = 2)  4! 
P( X = 2 | Y = 2) = = 0.6 2 0.3 2 0.10  0.2646 = 0.7347
P(Y = 2)  2!2!0! 
d.) E(X|Y=2)=0(0.0204)+1(0.2449)+2(0.7347) = 1.7142
5-31 a.), X has a binomial distribution with n = 3 and p = 0.01. Then, E(X) = 3(0.01) = 0.03
and V(X) = 3(0.01)(0.99) = 0.0297.

28. P( X | Y = 2)
b.) first find
P(Y = 2) = P( X = 1, Y = 2, Z = 0) + P( X = 0, Y = 2, Z = 1)
3! 3!
= 0.01(0.04) 2 0.95 0 + 0.010 (0.04) 2 0.951 = 0.0046
1!2!0! 0!2!1!
P( X = 0, Y = 2)  3! 
P( X = 0 | Y = 2) = = 0.010 0.04 2 0.951  0.004608 = 0.98958
P(Y = 2)  0!2!1! 
P( X = 1, Y = 2)  3! 
P( X = 1 | Y = 2) = = 0.0110.04 2 0.95 0  0.004608 = 0.01042
P(Y = 2)  1!2!1! 
E ( X | Y = 2) = 0(0.98958) + 1(0.01042) = 0.01042
V ( X | Y = 2) = E ( X 2 ) − ( E ( X )) 2 = 0.01042 − (0.01042) 2 = 0.01031
Section 5-3
3 2 3
5-35. a) P( X < 2, Y < 3) = 4
81 ∫ ∫ xydxdy =
0 0
4
81 (2) ∫ ydy = 81 (2)( 9 ) = 0.4444
0
4
2
b) P(X < 2.5) = P(X < 2.5, Y < 3) because the range of Y is from 0 to 3.
3 2. 5 3
P( X < 2.5, Y < 3) = 4
81 ∫ ∫ xydxdy =
0 0
4
81 (3.125) ∫ ydy = 81 (3.125) 9 = 0.6944
4
0
2
2. 5 3 2.5 2.5
y2
c) P(1 < Y < 2.5) = 4
81 ∫ ∫ xydxdy =
1 0
4
81 (4.5) ∫ ydy = 18
1
81 2
1
=0.5833
2.5 3 2.5
2
d) P( X > 1.8,1 < Y < 2.5) = 4
81 ∫ ∫ xydxdy =
1 1.8
4
81 (2.88) ∫ ydy =
1
4
81 (2.88) ( 2.52 −1) =0.3733
3 3 3 3
2
∫∫x ∫ 9 ydy = 4 y
2
e) E( X ) = 4
81
ydxdy = 4
81 9 2
=2
0 0 0 0
4 0 4
f) P( X < 0, Y < 4) = 4
81 ∫ ∫ xydxdy = 0∫ ydy = 0
0 0 0

29. 5-37.
3 x+2 3 x+2
y2
c ∫ ∫ ( x + y )dydx = ∫ xy + 2 dx
0 x 0 x
[ ]dx
3
( x+ 2)2
= ∫ x( x + 2) + 2 − x2 − x2
2
0
3
= c ∫ (4 x + 2)dx = 2 x 2 + 2 x [ ] 3
0
= 24c
0
Therefore, c = 1/24.
5-39. a) f X (x) is the integral of f XY ( x, y ) over the interval from x to x+2. That is,
x+2
1 1  y2
x+2
 x 1
f X ( x) =
24 ∫ ( x + y)dy =
x
24 

xy + 2
x
 = 6 + 12

for 0 < x < 3.
1
f XY (1, y )
(1+ y ) 1+ y
b) f Y 1 ( y) = f X (1) = 24
1 1 = for 1 < y < 3.
+
6 12
6
See the following graph,
y
2 f (y) defined over this line segment
Y|1
1
0 1 2 x
0
3
3 3
1+ y  1 1  y2 y3 
∫  6  61
dy = ∫ ( y + y )dy =   = 2.111
2
c) E(Y|X=1) = y +
1
6 2
 3 1

2
3 3
1+ y  1 1 y2 
d.) P (Y > 2 | X = 1) = ∫   dy = ∫ (1 + y )dy =  y +  =0.4167
2  6  61 6
 2 1

f XY ( x , 2 )
e.) f X 2 ( x) = fY ( 2) . Here f Y ( y) is determined by integrating over x. There are three regions of
integration. For 0< y≤2 the integration is from 0 to y. For 2< y≤3 the integration is from y-2 to y.
For 3 < y < 5 the integration is from y to 3. Because the condition is x=2, only the first integration is
y
1 1  x2 y
 y2
needed. fY ( y) = ∫
24 0
( x + y )dx =  2 + xy 0  =
24  
16 for 0 < y ≤ 2.

30. y
f X|2 (x) defined over this line segment
2
1
0 1 2 x
0
1
( x + 2)
24 x+2
Therefore, fY (2) = 1 / 4 and f X 2 ( x) = = for 0 < x < 2
1/ 4 6
5-43. Solve for c
∞ x ∞ ∞
c −2 x c −2 x
c ∫ ∫ e − 2 x −3 y dyd x = ∫ e (1 − e )d x = 3 ∫ e − e d x
−3 x −5 x
0 0
30 0
c 1 1 1
=  −  = c. c = 10
3  2 5  10
5-49. The graph of the range of (X, Y) is
y
5
4
3
2
1
0 1 2 3 4 x
1 x +1 4 x +1
∫ ∫ cdydx + ∫ ∫ cdydx = 1
0 0 1 x −1
1 4
= c ∫ ( x + 1)dx + 2c ∫ dx
0 1
= c + 6c = 7.5c = 1
3
2
Therefore, c = 1/7.5=2/15
5-51. a. )
x +1
1  x +1
f ( x) = ∫
0
7.5
dy =   for
 7.5 
0 < x < 1,
x +1
1  x + 1 − ( x − 1)  2
f ( x) =
x −1
∫7.5
dy =
 7.5
=
 7.5
for 1 < x < 4

31. b. )
f XY (1, y ) 1 / 7.5
f Y | X =1 ( y ) = = = 0.5
f X (1) 2 / 7.5
f Y | X =1 ( y ) = 0.5 for 0 < y < 2
2 2
y y2
c. ) E (Y | X = 1) = ∫ dy = =1
0
2 4 0
0.5 0.5
d.) P (Y < 0.5 | X = 1) = ∫ 0.5dy = 0.5 y = 0.25
0 0
5-53 a.) µ=3.2 λ=1/3.2
∞ ∞
−
x
−
y ∞
−
x
 − 352 
P ( X > 5 , Y > 5 ) = 10 . 24 ∫ ∫ e 3 .2 3 .2
dydx = 3 . 2 ∫ e 3 .2 e .  dx
 
5 5 5  
 −5   − 352 
=  e 3 .2

 e .

 = 0 . 0439

  
∞∞
−
x
−
y ∞
−
x
 − 3.2 
10
P( X > 10, Y > 10) = 10.24 ∫ ∫ e 3.2 3.2
dydx = 3.2 ∫ e 3.2  e dx
 
10 10 10  
 −
10
 −
10

= e

3.2  e

3.2  = 0.0019

  
b.) Let X denote the number of orders in a 5-minute interval. Then X is a Poisson
random variable with λ=5/3.2 = 1.5625.
e −1.5625 (1.5625) 2
P( X = 2) = = 0.256
21
2
For both systems, P ( X = 2) P ( X = 2) = 0.256 = 0.0655
c.) The joint probability distribution is not necessary because the two processes are
independent and we can just multiply the probabilities.
Section 5-4
0. 5 1 1 0.5 1 0.5 0.5
∫ ∫ ∫ (8 xyz)dzdydx = ∫ ∫ (4 xy)dydx = ∫ (2 x)dx = x
2
5-55. a) P( X < 0.5) = = 0.25
0 0 0 0 0 0 0
b)